Question

A solution contains $10mL0.1NNaOH$ and $10mL0.05N{H}_{2}S{O}_4,PH$ of this solution is

• A. less than 7
• B. 7
• C. zero
• D. greater than 7

Answer 1

Answer: (D)

$10mL$ of $0.1N(0.1M)NaOH=10\times 0.1=1$ millimole
$10mL$ of $0.05N(0.025M)H_{2}S{O}_4=10\times 0.025$
$=0.25$ millimole
$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+H_{2}O$
$1mol2mol$
$0.25$ millimole of $H_{2}S{O}_4$ will react with
$0.5$ millimole of $NaOH$.
$\therefore NaOH$ left $=1-0.5=0.5$ millimole
Volume of reaction mixture $=20mL$
Molarity of $NaOH$ in mixture solution
$=\frac{0.5}{20}=2.5\times 10^{-2}M$
$pOH=-\log \left[O{H}^{-}\right]=-\log \left(2.5\times 10^{-2}\right)$
$=1.60$
$pH=12.4$